SQL服务器：GROUP BY日期时间无蜱

Question

我有一个查询，返回所有条目，具有独特的datetime蜱：

select t.date, count(*) as Count 
    from table t 
    group by t.date having count(*) = 1 

我需要一个查询将返回独特datetime条目没有蜱。为了得到datetime无蜱我使用：

select CONVERT(VARCHAR(19), t.date, 120), count(*) as Count 
from table t 

所以我希望使用查询：

select CONVERT(VARCHAR(19), t.date, 120), count(*) as Count 
    from table t 
    group by CONVERT(VARCHAR(19), t.date, 120) having count(*) = 1 

但它抛出一个错误：

Column "table.date" is invalid in the ORDER BY clause because it is not contained in either an aggregate function or the GROUP BY clause.

你有什么我应该使用什么查询？

Answer 1

这工作得很好：

select convert(varchar(19), t.date, 120), count(*) as count 
from table t 
group by convert(varchar(19), t.date, 120) 
having count(*) = 1 
order by convert(varchar(19), t.date, 120) 

Answer 2

有每个组中多个可能datetime值，所以不是

ORDER BY t.date 

可以使用，例如

ORDER BY MIN(t.date) 

避免这种错误。

Answer 3

你必须ORDER BY在GROUP BY或表达汇聚

ORDER BY 
    CONVERT(VARCHAR(19), date, 120) 

例如：

SELECT 
    ShorterDateTime, COUNT(*) 
FROM 
    (
    select CONVERT(VARCHAR(19), date, 120) AS ShorterDateTime 
    from table 
    ) t 
GROUP BY 
    ShorterDateTime 
HAVING 
    COUNT(*) > 1 
ORDER BY 
    ShorterDateTime